1. (2opts). An Univeristy ATM handles on average 1000 transactions per day with

Question

1. (2opts). An Univeristy ATM handles on average 1000 transactions per day with a variance of 625 transactions. (a) What can be said about the probability that this ATM handles handles less than 900 transactions or more than 1100 transactions on any day? (h) What about the probability that it handles between 900 transactions and 1100 transact ions on any day? Hints: Since there is not enough information, the probabilities can be approximated by Chebyshev Inequalities. In (a) use P(|X?Mu| > = k sigma)

Explanation / Answer

Mean = u = 1000

Variance = 625

So, SD = sigma = sqrt(625) = 25

Now, between 900 and 1100 -->

Between 1000 - 4*25 and 1000 + 4*25, means we are finding the number of values lying within 4 standard deviations from the mean...

This indicates that k = 4

So, from the table, we know that 93.75% lies within 4 standard deviations from the mean...

This would mean that 93.75% lie within 900 and 1100

So, 100% - 93.75% = 6.25% lies outside the range (900 - 1100)

So, P(less than 900 or more than 1100) = 6.25% or 0.0625 ---> FIRST ANSWER

P(between 900 and 1100) = 93.75% = 0.9375 --> SECOND ANSWER

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---